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I know that the the set of real numbers has been proved uncountable by mathematicians, so my question is why this is wrong.

In countability arguments that I have seen the numbers are laid out in a grid with 1,2,3,4,... as rows and columns and the number at the nth row and mth column is n/m. So what if we have an extra dimension k, that represents the power, so at depth 3 the power is 1/3

so it becomes $ ({\frac{n}{m}})^{(\frac{1}{k})} $.

It should be possible to visit every position in the cube by a similar procedure akin to the one used for 2 dimensional grid.

Any non fractional power will give another rational number so only 1/k is actually necessary. While this only appears to work for the solution to a polynomial, such solutions are not rational numbers. Which would make some sets of irrational numbers countable, but that doesn't quite make sense to me.

So therefore there must be something wrong with my ideas but I don't know what it is.

Sam
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    "Any non fractional power will give another rational number so only 1/k is actually necessary." -- this is certainly not true. Can you show that $2^\pi$ is rational? – tomasz Jul 06 '14 at 21:05
  • sorry my question is not clear, a rational number n/m to the power of 2/3 is the same as $n^2$/$m^2$ to the power of 1/3 – Sam Jul 06 '14 at 21:08
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    Actually, everything is right. Or at least was right at the time when you wrote it. Now all of the maths is still right, but the part where you state your confusion hopefully isn't anymore ;) – Carsten S Jul 06 '14 at 21:18
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    Some part of what you are saying seems to be that you found that the set of so-called "algebraic numbers", which include the irrational numbers that are zeroes of polynomials with integer coefficients, is countable. That is correct. It is the transcendental irrationals (which constitute "almost all" of the real numbers) that are uncountable. – colormegone Jul 06 '14 at 21:38
  • I think the analogue to the infinite square representing the rationals for the reals is an infinite set of infinite squares. – rlms Jul 07 '14 at 10:41

3 Answers3

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As you point out, every number appearing in your cube is a root of a polynomial with rational coefficients, so it is an algebraic number. And the set of algebraic numbers over $\mathbb{Q}$ is countable (for example, see Prove that the set of all algebraic numbers is countable).

Community
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rogerl
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  • It should be noted that the cube does not enumerate all algebraic numbers; rather it only goes through the rationals, and all possible roots of the rationals - namely, the subset consisting of all roots to polynomials of the form $x^n-q,;q\in\mathbb{Q}$. Most algebraic numbers cannot be expressed in closed form this way. –  Jul 07 '14 at 02:43
  • @NotNotLogical Agreed. I should have pointed out that the enumerated set is a subset of the set of algebraic numbers. – rogerl Jul 07 '14 at 03:42
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You have only found a countable set which includes SOME irrational numbers. How do you know ALL of them can be written in that way?

What I am saying is, there are some irrational numbers ($\pi$ or any other transcendental number, for example), which cannot be written as $q^{1/k}$ for some rational number $q$ and integer $k$.

StrangerLoop
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There is nothing wrong with some sets of irrational numbers being countable. For example, $\{\sqrt 2\}$ is a countable set of irrational numbers, as is the set of all (irrational) algebraic numbers (that is, roots of polynomials with rational coefficients; this includes all the numbers you have listed), or the singleton of any real number whatsoever (or even any singleton at all, for that matter).

But that is beside the point. The point is, the set of all real numbers is not countable, as Cantor's diagonal argument shows. I'm not going to cite this argument here, as it is readily found just about anywhere (like Wikipedia).

tomasz
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